Question

The major product obtained in the following reaction is:

(A)
(B)
(C)
(D)

Answer 1

The base causes the abstraction of the proton, and there are two carbonyl groups present which will undergo aldol condensation reaction within the molecule.

Complete step by step answer:
In the given compound, there is a presence of two carbonyl groups at both the ends. That is, an aldehyde and a ketone. Both the carbonyl groups have hydrogens on the $\alpha$- carbon adjacent to it.
So, in presence of the strong base of sodium hydroxide, it undergoes the intramolecular cross aldol condensation as follows:
- The hydroxide ion of the base deprotonates the carbonyl group, abstracting the more acidic $\alpha$- hydrogen. Since, the $\alpha$- hydrogen near the aldehyde group due to the presence of the methyl group over the $\alpha$- carbon, increases its electron-density. Thus, making the $\alpha$- hydrogen less acidic.
Whereas the methylene carbon adjacent to the ketone group, forming a secondary carbanion is more acidic. So, the sodium hydroxide base, abstracts hydrogen from the methylene carbon, forming an enolate ion.

The negative charge disperses over the carbonyl group, forming a double bond and a negative charge over oxygen.
- The enolate ion attacks on the partially positive charge on the carbonyl carbon of the aldehyde group. Thus, forming a 5-membered ring and a $\beta -hydroxyketone$
- On further heating, it loses a water molecule, forming an $\text{ }\\!\\!\alpha\\!\\!\text{ , }\\!\\!\beta\\!\\!\text{ - unsaturated ketone}$ compound.

Therefore, the major product obtained in the intramolecular cross aldol condensation reaction is option (D).

Note: It is an intramolecular reaction, as both the carbonyl groups, that is the aldehyde and ketone group are present within the molecule and the cross-aldol condensation takes place. On either side of the ketone, the $\alpha$- hydrogen of the methylene carbon is more acidic than the methyl group.